Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{-2r - 2}{r + 7} \times \dfrac{-6r - 42}{r^2 - 1} $
First factor the quadratic. $q = \dfrac{-2r - 2}{r + 7} \times \dfrac{-6r - 42}{(r + 1)(r - 1)} $ Then factor out any other terms. $q = \dfrac{-2(r + 1)}{r + 7} \times \dfrac{-6(r + 7)}{(r + 1)(r - 1)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ -2(r + 1) \times -6(r + 7) } { (r + 7) \times (r + 1)(r - 1) } $ $q = \dfrac{ 12(r + 1)(r + 7)}{ (r + 7)(r + 1)(r - 1)} $ Notice that $(r + 7)$ and $(r + 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ 12\cancel{(r + 1)}(r + 7)}{ (r + 7)\cancel{(r + 1)}(r - 1)} $ We are dividing by $r + 1$ , so $r + 1 \neq 0$ Therefore, $r \neq -1$ $q = \dfrac{ 12\cancel{(r + 1)}\cancel{(r + 7)}}{ \cancel{(r + 7)}\cancel{(r + 1)}(r - 1)} $ We are dividing by $r + 7$ , so $r + 7 \neq 0$ Therefore, $r \neq -7$ $q = \dfrac{12}{r - 1} ; \space r \neq -1 ; \space r \neq -7 $